If displacement $x$ and velocity $v$ are related as $4v^2 = 16\, -\, x^2$ in a $SHM$ . Then time period of given $SHM$ is (consider $SI\,units$ )
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$4 v^{2}=16-x^{2} \Rightarrow \frac{x^{2}}{16}+\frac{v^{2}}{4}=1$
relation between displacement and velocity is
$\frac{x^{2}}{A^{2}}+\frac{v^{2}}{(A \omega)^{2}}=1$
$\therefore A \omega=2$ and $A=4$
$\omega=\frac{1}{2} \quad \Rightarrow \mathrm{T}=4 \pi$
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